Comparing Olympus 4/3lenses to FX "Full Frame" offerings

Started 7 months ago | Discussions thread
bobn2
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Re: Oh?
In reply to Tiger1, 7 months ago

Tiger1 wrote:

Great Bustard wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Nikon 300 F4 to Zuiko ed 150 F2 to canon 300 F4

No this is where we disagree. Olympus 150mm F2 is the same as a 300mm F2 FF lens.

It is not. It has neither the same focal length nor aperture diameter.

Of course it's not literally!

Then why did you say it was. You said 'Olympus 150mm F2 is the same as a 300mm F2 FF lens'. It isn't.

Are you that daft?

He isn't, I'm beginning to wonder about you.

I'm talking of field of view and magnification and amount of light hitting the sensor per unit area. In those respects it is!!!!!!!

In some more important aspects, like DOF and total light hitting the sensor, it isn't. So what you really meant to say is 'Olympus 150mm F2 is the same as a 300mm F2 FF lens except fro some of the more important things'. After all, what do you use the aperture control for? To control DOF and to control the amount of light hitting the sensor. In those two aspects, the two are not alike.

I'm sorry but now you're getting pedantic. Look through this entire site and what manufacturers write next to their lenses. By convention it is written F2 even though in reality it is f/2. You are stating something obvious that has nothing to do with the argument.

If you like f/2 = f/2. There!

The point isn't 'pedantry' - it's about understanding what the aperture is. People writing 'F2' seem to think that aperture is measured in 'effs' and that there are two of them in that lens, this isn't right. What it means is that the aperture of the lens is its focal length divided by two.

FF sensors are not "less noisy". Instead, what happens is that the larger aperture diameter of FF lenses projects more light onto the sensor for a given shutter speed, and it is this greater amount of light falling on the sensor, not the sensor itself, that makes FF less noisy.

Gee whiz. Give up!

You should, it makes more sense that trying to lay down the law about things that you don't know about.

FF sensors using a certain design/technology are less noisy than smaller format sensors with equivalent photosite count and design/technology BECAUSE their photosites are larger and therefore collect more photons and so need less amplification of signal (which introduces noise).

This is not true, any of it. Think, a competent amplifier has a SNR of 80dB or more, but with sensors we are talking SNRs of about 40dB. The amplifier noise is 40dB below the photon noise. The idea that small pixels are more noisy because they need 'more amplification' is a myth. It's also untrue, small pixels don't need more amplification because they have higher conversion gain.

Basic physics!

Best to avoid quoiting physics if you don't know it. Physics says that the predominant noise in digital imaging is photon shot noise, which depends on how many photons make up the image.

The light hitting the sensor per unit are is THE SAME. Because the area of the sensor is larger the amount of light IN TOTAL is larger as it's unit area is larger. The larger aperture allows more light spread not more intense light!!!!

You state this as though it's a great truth. It's a simple fact, which no-one has denied. The question is, what is the impact of having the same light per unit area on the final image. And the truth is, nothing at all, so long as the processing is appropriate to the capture made.

Imagine you are peering through a hole at the sun. It's bright. Now that hole becomes really small. It would seem the light intensity has fallen, right? No. it is the same. Just spread across a smaller area. If you don't think so, imagine now that you shrink yourself into an ant. That hole is now relatively large again and the sun seems just as bright. That is what is happening with the format changes. Understand now?

What do you understand to be the importance of light intensity, as opposed to the total light?

Then you can argue that the quality of both shots would be roughly equal.

There is more to a photo than the quantity of light that falls on the sensor, but the amount of light falling on the sensor, coupled with the sensor efficiency, does determine the noise in the photo.

I know buddy. I'm talking about quality of image creation via the sensor. Thanks.

You clearly are not talking about that. since you seem to be ignorant of the basic physics.

It is important to note that the DOF will be not be equivalent and I have always agreed on that.

What you, and 99% of the people on this forum, fail to understand is the relationship between the relative aperture (f-ratio), the virtual aperture (entrance pupil), the amount of light per area falling on the sensor (exposure), and the total amount of light collected by the sensor, and how these quantities relate to both noise and DOF.

I understand. You don't.

You got that precisely the wrong way round.

You, and those like you, would do well to make an effort to understand these things, rather than simply chanting "F2=F2!!!" with religious fervor, having no real understanding about how and why things are the way they are.

Mate. Go back to school.

Hopefully, not the same school that you went to.

Here, let me make it plain to you. Let's say we have an absolutely perfect 2x TC (that is, a 2X TC that is completely free of aberrations) and mounted it behind the 150 / 2. The 150 / 2 is now a 300 / 4, right?

Yes obviously.

Now, if we put that 300 / 4 in front of a FF sensor, the image it will record will be all but identical to the image recorded by the bare lens in front of a 4/3 sensor. Of course, I'm assuming you know how a TC works, so...

No it won't mate. The TC optically magnifies the image leading to a four fold loss of light.

Where do you think the lost light goes...?

The 4/3 sensor "magnifies" the image because its like putting the 150mm F2 lens in front of a FF sensor then taking the middle part of the image and "digitally" magnifying it.

So what happens to the light that is outside the 'middle part'?

The amount of light hitting the sensor is the same for both per unit area. How could it be any different?

Is that really true? Do you really believe that the light hitting the sensor per unit area is the same with a TC as without it? If so, how does the optical magnification work?

THIS IS SO OBVIOUS IT'S NOT FUNNY.......

What seems 'obvious' is often so because it's just plain wrong.

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Bob

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